$S$ and $T$ are points on sides $PR$ and $QR$ of $∆PQR$ such that $\angle P = \angle RTS$. Show that $∆RPQ \sim ∆RTS$.
Given:
$S$ and $T$ are points on sides $PR$ and $QR$ of $∆PQR$ such that $\angle P = \angle RTS$.
To do:
We have to show that $∆RPQ \sim ∆RTS$.
Solution:
In $\triangle RPQ$ and $\triangle RTS$,
$\angle P=\angle RTS$
$\angle P=\angle R$
Therefore, by AA criterion,
$\Delta RPQ \sim \Delta RTS$
Hence proved.
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